Function

Anon

hi people I have a little question but it's an important one for me.

$name="frank";
echo "$name";-------This does display Frank on the screen.----

function show(){
echo "$name";----This DOESN'T display Frank on the screen.----

}

function show2(){
$name="Henk";
echo "$name";-------This does display Henk on the screen.----
}

so please tell me how to get a string or integer in a function.

toma42

so please tell me how to get a string or integer in a function

I would start by reading the manual...try the section near the very beginning about declaring functions.

Anon

if you go like this:

function show($name) {
echo $name;
}

and then later on go like this

show(Frank)

it will echo frank

Frank wrote:

function show(){
echo "$name";----This DOESN'T display Frank on the screen.----

}

what you did there wont work because nothing is set to $name

Anon

That's not the problem it should display the $name which was set before the function. I know how to display things FROM a function but not how to display a thing UT OF a function IN a function

Anon

I did, I read the whole fucking manual

Anon

Either pass in $name

function show($name){
echo $name;
}

or tell it that it's a global. And you don't need to put quotes around $name in the echo statement. And RTFM on variable scope. http://www.php.net/manual/en/language.variables.scope.php

function show($name){
global $name;
echo $name;
}

or -

function show($name){
echo $GLOBALS["name"];
}

Anon

Not really possible without a database. It's hard and uneffiecent to do when you do have a database and extremly extremly uneffiecnt without a database, but ofcourse, it's possible and takes some brianstorming 🙂

Anon

Wrong Thread, sorry.

toma42

er, including this part (near the beginning)?

http://www.php.net/manual/en/functions.arguments.php

Anon

That's not the problem it should display the $name which was set before the function. <<

Just because you set a variable somewhere in your code doesn't mean a thing to your function. It knows nothing about it!

Where in your example did you pass Show() your $name variable? I didn't see it.

You've either got to pass it the variable, $name, or, as the others have shown you, declare $name as a global. That's poor programming practice, but it will work.

And note that PHP globals take extra handling. Unlike C or C++, they must also be declared inside the body of the function (again, as was shown).

10-4 ?